Problem: Divide the following complex numbers. $ \dfrac{-19+4i}{-2+5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2-5i}$ $ \dfrac{-19+4i}{-2+5i} = \dfrac{-19+4i}{-2+5i} \cdot \dfrac{{-2-5i}}{{-2-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-19+4i) \cdot (-2-5i)} {(-2+5i) \cdot (-2-5i)} = \dfrac{(-19+4i) \cdot (-2-5i)} {(-2)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-19+4i) \cdot (-2-5i)} {(-2)^2 - (5i)^2} = $ $ \dfrac{(-19+4i) \cdot (-2-5i)} {4 + 25} = $ $ \dfrac{(-19+4i) \cdot (-2-5i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-19+4i}) \cdot ({-2-5i})} {29} = $ $ \dfrac{{-19} \cdot {(-2)} + {4} \cdot {(-2) i} + {-19} \cdot {-5 i} + {4} \cdot {-5 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{38 - 8i + 95i - 20 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{38 - 8i + 95i + 20} {29} = \dfrac{58 + 87i} {29} = 2+3i $